TOPIC: Frame of Reference, Distance, Displacement, Average & Instantaneous Velocity/Acceleration
IIT-JEE MAINS PYQ Arena — Set 17
Frame of Reference, Distance, Displacement & Avg/Instantaneous Kinematics
Set 17 • Kinematics III
Instantaneous Acceleration from v = b x−2n
JEE Main 2024 (Jan Shift 1)
⏱ JEE Standard Pace: 120s
+4 / -1 Scale
Q1. The instantaneous velocity of a particle moving along the positive x-axis is given by the relation v = b x−2n, where b and n are constants and x is the position. What is the instantaneous acceleration of the particle as a function of x?
✔ Spot On! Chain Rule a = v (dv/dx) Mastered! +4 Marks
✖ Differentiate dv/dx = −2n b x−2n−1, then multiply by v = b x−2n to add the exponents!
Concept & Solution:• Spatial derivative of velocity: dv/dx = −2n b x−2n − 1.
• Instantaneous acceleration: a = v (dv/dx) = [b x−2n] × [−2n b x−2n − 1] = −2n b² x−4n − 1.
Question 1 of 15
Multi-Stage Average Speed (1/3rd & 2/3rd Distance)
JEE Main 2024 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q2. A train covers the first one-third (1/3) of the distance between two stations at a constant speed of 20 km/h, and the remaining two-thirds (2/3) of the distance at a constant speed of 60 km/h. What is the average speed of the train for the entire journey?
✔ Weighted Harmonic Mean Speed Nailed! +4 Marks
✖ Total time t = (d/3)/20 + (2d/3)/60 = d/60 + d/90 = 5d/180, so v_avg = 180/5 = 36 km/h!
Concept & Solution:• Let total distance be d.
• Time for first 1/3rd: t₁ = (d / 3) / 20 = d / 60.
• Time for remaining 2/3rds: t₂ = (2d / 3) / 60 = 2d / 180 = d / 90.
• Average Speed = d / (t₁ + t₂) = 1 / (1/60 + 1/90) = 180 / (3 + 2) = 36 km/h.
Frame of Reference (Balloon Upward Coin Release)
JEE Main 2023 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q3. A hot-air balloon is rising vertically upwards with a constant velocity of 10 m/s. When the balloon is at a height of 40 m above the ground, a stone is gently released from it. Taking g = 10 m/s², what are the Total Distance travelled byнти the stone in the air and the magnitude of its Net Displacement before hitting the ground (as seen from the ground frame)?
✔ Upward Inertia + Turning Point Distance Mastered! +4 Marks
✖ In the ground frame, the stone starts with u = +10 m/s upwards, rises by h = u²/(2g) = 5 m, and then falls 45 m to the ground!
Concept & Solution:• In the ground frame, the stone inherits the upward velocity of the balloon: u = +10 m/s.
• Additional height risen before stopping: Δh = u² / (2g) = 10² / (2 × 10) = 5 m.
• Total Distance = 5 m (up) + (5 + 40) m (down) = 5 + 45 = 50 m.
• Net Displacement = |y_final − y_initial| = 40 m downwards.
Separation of Variables (a = −k v³)
JEE Main 2023 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q4. A high-speed projectile enters a resistive medium at x = 0 with initial speed v₀ and experiences a deceleration given by a = −k v³, where k is a positive constant. What is the instantaneous velocity v(x) of 운동 the projectile as a function of penetration distance x?
✔ Differential Reduction a = v(dv/dx) = −kv³ Nailed! +4 Marks
✖ Substitute a = v(dv/dx) = −k v³ ⇒ dv/v² = −k dx, and integrate from v₀ to v!
Concept & Solution:• Using a = v (dv/dx) = −k v³ ⇒ dv / v² = −k dx.
• Integrating from (x = 0, v = v₀) to (x, v):
• [ −1 / v ]v₀v = −k x ⇒ 1/v − 1/v₀ = k x ⇒ 1/v = (1 + k v₀ x) / v₀.
• Therefore, v(x) = v₀ / (1 + k v₀ x).
Frame of Reference (Two Trains Crossing in Opposite Directions)
JEE Main 2022 (June Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q5. Train A of length 120 m is moving North at a constant speed of 20 m/s, while Train B of length 180 m is moving South on a parallel track at a constant speed of 30 m/s. How much time does it take for the two trains to completely cross each other, and what is the time taken by a single passenger sitting in Train A to cross Train B?
✔ Train-vs-Train and Point-Passenger-vs-Train Distinction Nailed! +4 Marks
✖ A passenger in Train A is a point observer, so the relative distance to cross Train B is only the length of Train B (180 m)!
Concept & Solution:• Relative speed of approach: v_rel = 20 + 30 = 50 m/s.
• Time for two trains to cross each other: t₁ = (L_A + L_B) / v_rel = (120 + 180) / 50 = 300 / 50 = 6.0 s.
• Time for a point passenger in Train A to cross Train B: t₂ = L_B / v_rel = 180 / 50 = 3.6 s.
2D Instantaneous Angle Between Velocity & Acceleration
JEE Main 2024 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q6. The position vector of a particle moving in the xy-plane is given by r(t) = (3t î + 2t² ĵ) m, where t is in seconds. What is the angle θ between its instantaneous velocity vector v and instantaneous acceleration vector a at t = 1 s?
✔ Dot Product cos θ = (v · a) / (|v||a|) Nailed! +4 Marks
✖ At t = 1 s, v = 3î + 4ĵ and a = 4ĵ, so cos θ = 16 / (5 × 4) = 4/5!
Concept & Solution:• Instantaneous Velocity: v(t) = dr/dt = 3 î + 4t ĵ ⇒ at t = 1 s, v(1) = 3 î + 4 ĵ m/s (|v| = 5 m/s).
• Instantaneous Acceleration: a(t) = dv/dt = 4 ĵ m/s² (|a| = 4 m/s²).
• cos θ = (v · a) / (|v| |a|) = (0 + 16) / (5 × 4) = 4 / 5 ⇒ θ = cos−1(4 / 5) ≈ 37°.
Harmonic Oscillation Distance vs. Displacement
JEE Main 2023 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q7. A particle oscillates along the x-axis according to the equation x(t) = 5 sin(π t / 2) cm, where t is in seconds. What are the Net Displacement and the Total Distance travelled by the particle from t = 0 s to t = 3 s?
✔ Quarter-Period Amplitude Tracking Mastered! +4 Marks
✖ Period T = 2π / (π/2) = 4 s, so in 3 s (3/4th of a cycle) the particle covers 3 × A = 15 cm of distance!
Concept & Solution:• Time period T = 2π / (π / 2) = 4 s, and Amplitude A = 5 cm.
• Position at t = 0 s: x(0) = 0; Position at t = 3 s: x(3) = 5 sin(3π / 2) = −5 cm.
• Net Displacement = x(3) − x(0) = −5 cm.
• In 3 seconds (three quarter-periods), the particle moves 0 → +5 cm → 0 → −5 cm: Total Distance = 3 A = 15 cm.
Frame of Reference (River-Boat Minimum Time vs. Drift)
JEE Main 2024 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q8. A river of width W = 400 m is flowing East at a speed of 3 m/s. A motorboat that can move at 4 m/s in still water is steered perpendicular to the river banks (due North) so as to cross the river in the minimum possible time. What are the time taken to cross the river and the downstream drift (x) of the boat when it reaches the opposite bank?
✔ Independence of Orthogonal Velocity Components! +4 Marks
✖ Crossing time depends ONLY on the perpendicular component (400 / 4 = 100 s), and drift = v_river × t = 3 × 100 = 300 m!
Concept & Solution:• Velocity component perpendicular to banks: v_y = 4 m/s.
• Minimum Crossing Time: t_min = W / v_y = 400 m / 4 m/s = 100 s.
• Downstream Drift along the river flow: x = v_river × t_min = 3 m/s × 100 s = 300 m (Total net displacement = √(400² + 300²) = 500 m).
Average Acceleration on a Semicircular Track
JEE Main 2022 (July Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q9. A particle moves with a constant speed v = 10 m/s along a circular path of radius R = 5 m. What is the magnitude of the average acceleration of the particle during the time it completes a half-revolution (semicircle)?
✔ Semicircle |Δv| = 2v & a_avg = 2v²/(πR) Mastered! +4 Marks
✖ Instantaneous centripetal acceleration is v²/R = 20 m/s², but Average Acceleration over a semicircle is |Δv|/Δt = 2v / (πR/v) = 40/π m/s²!
Concept & Solution:• Over a half-revolution (180° turn), velocity reverses direction, so |Δv| = v − (−v) = 2v = 20 m/s.
• Time taken for half-revolution: Δt = (π R) / v = (5 π) / 10 = (π / 2) s.
• Magnitude of Average Acceleration: |a_avg| = |Δv| / Δt = 20 / (π / 2) = (40 / π) m/s².
Instantaneous Velocity from Quadratic t(x)
JEE Main 2021 (Mar Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q10. The time t (in seconds) and position x (in meters) of a particle moving along the positive x-axis satisfy the equation t = x² + 2x. What is the instantaneous velocity and instantaneous acceleration of the particle when it passes through x = 1 m?
✔ Numerical Application of a = −2α v³ Nailed! +4 Marks
✖ At x = 1 m, dt/dx = 2(1) + 2 = 4 ⇒ v = 1/4 = 0.25 m/s, and a = −2(1) v³ = −2(1/64) = −1/32 m/s²!
Concept & Solution:• Differentiating with respect to x: dt/dx = 2x + 2.
• At x = 1 m, dt/dx = 4 s/m ⇒ Instantaneous Velocity v = dx/dt = 1/4 = 0.25 m/s.
• Using a = −2 α v³ (where α = 1): a = −2 × (1/4)³ = −2 / 64 = −1/32 m/s².
Frame of Reference (Wind-Cyclist Apparent Direction)
JEE Main 2023 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q11. To a cyclist riding due East at 15 km/h, the wind appears to blow directly from the North. When the cyclist doubles his speed to 30 km/h due East, the wind appears to blow from the North-East (at 45°). What is the true velocity vector of the wind with respect to the ground (taking East as +î and North as +ĵ)?
✔ Classic Two-State Relative Velocity System Solved! +4 Marks
✖ When v_c = 15 î, v_wc has zero x-component, so v_wx = +15 km/h!
Concept & Solution:• Let true wind velocity be v_w = v_x î + v_y ĵ.
• Case 1 (v_c1 = 15 î): v_wc1 = (v_x − 15) î + v_y ĵ blows from North (towards −ĵ), so v_x − 15 = 0 ⇒ v_x = +15 km/h.
• Case 2 (v_c2 = 30 î): v_wc2 = (15 − 30) î + v_y ĵ = −15 î + v_y ĵ blows from North-East (at 45° towards South-West), so v_y = −15 km/h.
• True Wind Velocity = (15 î − 15 ĵ) km/h (magnitude = 15√2 km/h).
2D Instantaneous Speed Minimum
JEE Main 2024 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q12. A particle moves in the xy-plane such that its velocity vector at time t is given by v(t) = (t − 2) î + (t + 2) ĵ m/s. What is the minimum instantaneous speed attained by the particle for t ≥ 0, and at what time t does it occur?
✔ Speed Squared Function v²(t) = 2t² + 8 Minimized! +4 Marks
✖ Write v²(t) = (t − 2)² + (t + 2)² = 2t² + 8, which is minimum at t = 0 s (giving v_min = √8 = 2√2 m/s)!
Concept & Solution:• Instantaneous Speed squared: v²(t) = v_x² + v_y² = (t − 2)² + (t + 2)² = 2 t² + 8.
• Clearly, 2t² + 8 is minimum at t = 0 s.
• Therefore, v_min = √8 = 2√2 m/s at t = 0 s (where v · a = 0 also holds).
Parabolic Velocity Function & Distance with Reversal
JEE Main 2023 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q13. The velocity of a particle moving along a straight line is given by v(t) = (6 − 2t) m/s. What is the Total Distance travelled by the particle during the time interval from t = 0 s to t = 5 s, and what is its Average Speed over this 5 s interval?
✔ Absolute Area Under v-t Graph Split at t = 3 s! +4 Marks
✖ Velocity becomes zero at t = 3 s! Add the positive area from [0, 3] (9 m) and the magnitude of negative area from [3, 5] (4 m)!
Concept & Solution:• Velocity v(t) = 6 − 2t = 0 at t = 3 s (turning point).
• Distance from t = 0 to t = 3 s: d₁ = ∫₀³ (6 − 2t) dt = (1/2) × 3 × 6 = 9 m.
• Distance from t = 3 to t = 5 s: d₂ = |∫₃⁵ (6 − 2t) dt| = (1/2) × 2 × 4 = 4 m.
• Total Distance = 9 + 4 = 13 m ⇒ Average Speed = 13 m / 5 s = 2.6 m/s (while Net Displacement = 9 − 4 = 5 m).
Frame of Reference (Closest Approach of Two Ships)
JEE Main 2022 (July Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q14. At t = 0, Ship A is at the origin moving due East at 10 km/h, while Ship B is located 10 km due East of Ship A and is moving due North at 10 km/h. Working in the reference frame of Ship A, what is the minimum distance (distance of closest approach) between the two ships, and at what time does it occur?
✔ Relative Frame Closest-Approach Geometry Mastered! +4 Marks
✖ In Ship A’s frame, v_BA = −10 î + 10 ĵ km/h (at 45°), so d_min = 10 sin 45° = 5√2 km and t = (5√2) / (10√2) = 0.5 h!
Concept & Solution:• In the reference frame of Ship A (fixing A at origin):
• Initial position of B: r_BA(0) = 10 î km.
• Relative velocity of B: v_BA = v_B − v_A = −10 î + 10 ĵ km/h (speed |v_BA| = 10√2 km/h, directed at 45° North-West).
• Perpendicular distance from origin A to the relative line of motion: d_min = 10 sin 45° = 5√2 km.
• Time to reach closest approach: t = (10 cos 45°) / |v_BA| = (5√2) / (10√2) = 0.5 h (30 min).
SHM Relation Between Position & Acceleration
JEE Main 2024 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q15. The velocity v of a particle moving along the x-axis satisfies the equation v² = 36 − 9 x² (in SI units). What is the magnitude of the maximum acceleration of the particle during its motion?
✔ 255th Question Milestone Conquered! +4 Marks
✖ Differentiate v² = 36 − 9x² to get a = −9x, and at extreme position (v = 0 ⇒ x = ±2 m), |a_max| = 9 × 2 = 18 m/s²!
Concept & Solution:• Differentiating v² = 36 − 9 x² with respect to x:
• 2v (dv/dx) = −18 x ⇒ a = v (dv/dx) = −9 x.
• Maximum displacement occurs where v = 0: 36 − 9 x_max² = 0 ⇒ x_max² = 4 ⇒ x_max = 2 m.
• Therefore, Maximum Acceleration |a_max| = 9 × 2 = 18 m/s².
Set 17 — 255-PYQ Milestone Completed!
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